2023每日刷题(二十三)
Leetcode—LCR 044.在每个树行中找最大值
DFS实现代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
#define MAX(a, b) ((a > b ? (a) : (b)))
#define MAXSIZE 10003
void dfs(int *res, int cur, int *pos, struct TreeNode* root) {
if(*pos == cur) {
res[(*pos)++] = root->val;
} else {
res[cur] = MAX(res[cur], root->val);
}
if(root->left) {
dfs(res, cur + 1, pos, root->left);
}
if(root->right) {
dfs(res, cur + 1, pos, root->right);
}
}
int* largestValues(struct TreeNode* root, int* returnSize) {
*returnSize = 0;
if(root == NULL) {
return NULL;
}
int *res = (int *)malloc(sizeof(int) * MAXSIZE);
dfs(res, 0, returnSize, root);
return res;
}
运行结果
BFS实现代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
#define MAXSIZE 10003
#define MAX(a, b) ((a > b) ? (a) : (b))
int* largestValues(struct TreeNode* root, int* returnSize){
struct TreeNode **queue = (struct TreeNode **)malloc(sizeof(struct TreeNode*)*MAXSIZE);
*returnSize = 0;
int *res = (int *)malloc(sizeof(int)*MAXSIZE);
if(root == NULL) {
return NULL;
}
int pos = 0;
int front = 0, rear = 0;
int len = 0;
queue[rear++] = root;
while(front != rear) {
len = rear - front;
int maxVal = INT_MIN;
while(len > 0) {
len--;
struct TreeNode *tmp = queue[front++];
maxVal = MAX(maxVal, tmp->val);
if(tmp->left) {
queue[rear++] = tmp->left;
}
if(tmp->right) {
queue[rear++] = tmp->right;
}
}
res[pos++] = maxVal;
}
*returnSize = pos;
free(queue);
return res;
}
运行结果
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